A ladder leans against a frictionless wall. If the coefficient of friction with the ground is \(\mu\), what is the smallest angle that the ladder can make with the ground and not slip?

Solution:

Let the ladder have mass m and length \(l\). We have three unknown forces: (i) the frictional force F (ii) the normal forces \(N_1\) and \(N_2\). And we fortunately have three equations that will allow us to solve for these three forces: \(\Sigma F_{vert}=0\), \(\Sigma F_{horiz}\) and \(\tau=0\). Now, from the figure we can see that \(N_1=mg\).And then when we look at the horizontal forces, we see that \(N_2=F\).

We will now use \(\tau = 0\) to find \(N_2\) (or F).

So the number of unknowns are reduced frook three to one. There are two forces acting at the bottom end of the ladder. Balancing the torques due to gravity and \(N_2\), we have $$N_2lsin\theta=mg\Bigr(\frac{l}{2}\Bigr)cos\theta$$

( The factor 1/2 comes into play because the ladder behaves like a point mass located halfway up)

$$\Rightarrow N_2=\frac{mg}{2tan\theta}$$ This is also the value of the frictional force F. The condition $$ F\mu N_1=\mu mg$$